# Joan Lindsay Orr

$$\newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\H}{\mathcal{H}} \newcommand{\e}{\epsilon} \newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}}$$

$$\newcommand{\d}{\partial}$$

# Lagrangian and Hamiltonian Mechanics

## Lagrangians

Lagrangian mechanics improves on the formulation of Newtonian mechanics because it

• avoids the necessity of consideration of all the forces in a system, which can be very complicated (e.g. double pendulum) and

• expresses everything in terms of generalized coordinates, which consist of spatial coordinates $$q_i$$ and their time derivatives $$\dot{q}_i$$. Generalized coordinates can be chosen to reflect the properties of the system under consideration. For example for a simple pendulum use $$(\theta,\dot{\theta})$$ and for a double pendulum use $$(\theta,\phi, \dot{\theta},\dot{\phi})$$.

Assume the external force acting on the system can be expressed $$\vec{F}=\nabla{V}$$, where $$V$$ is the potential energy. The kinetic energy is $T = \frac{1}{2}m\frac{d\vec{r}}{dt}\cdot\frac{d\vec{r}}{dt}$ Then define the Lagrangian of the system to be $$L=T-V$$.

Choose spatial coordinates $$q_i$$ to describe the physical state of the system and introduce corresponding new variables $$\dot{q_i}$$ which will represent $$dq_i/dt$$. Express $$T$$ and $$V$$ in terms of these new variables so that $$L = L(q_1, q_2, \ldots\, \dot{q_1}, \dot{q_2}, \ldots)$$. Regarding the solution path as a function $$\mathbf{q}(t) = (q_i(t))$$, The action is $A = \int_{t_0}^{t_1} L(\mathbf{q}(t), \dot{\mathbf{q}}(t)) dt$ and the Principle of Stationary Action says that the solution path is found at a stationary point of the action. By the Euler-Lagrange formula, the stationary value satisfies Lagrange's equations: $\frac{\d L}{\d q_i} = \frac{d}{dt}\frac{\d L}{\d \dot{q_i}}$

(A variational argument based on Newton's second law (i.e. $$\vec{F}\cdot\delta\vec{r}=m d^2\vec{r}/dt^2\cdot\delta\vec{r}$$) can also derive Lagrange's Equations.)

For a simple pendulum, $$V=-mgl\cos\theta$$ and $$T=\frac{1}{2}ml^2\dot\theta^2$$. Thus $L(\theta, \dot\theta,t) = \frac{1}{2}ml^2\dot\theta^2 +mlg\cos\theta$ and so $\begin{split} \frac{\d L}{\d\theta} &= -mlg\sin\theta \\ \frac{\d L}{\d\dot\theta} &= ml^2\dot\theta\\ \frac{d}{dt}\frac{\d L}{\d\dot\theta} &= ml^2\ddot\theta \end{split}$ Thus Lagrange's Equation reduces to the familiar equation of motion $\ddot\theta = -\frac{g}{l}\sin\theta$

## Legendre Transform

Let $$f$$ be a convex function $$\RR\rightarrow\RR$$ (i.e. $$\ddot{f}(x)>0$$). Say that $$f^\star$$ is the Legendre Transform of $$f$$ if $f^\star(p) + f(x) = px \text{ where } p = f'(x)$ Equivalently, $$f'$$ and $$(f^\star)'$$ are mutual inverses of each other. Note the connection with tangent lines; the tangent line to $$y=f(x)$$ at $$x=x_0$$ is $$y=mx+b$$ where $m = f'(x_0) = p_0$ and $b=y_0-mx_0=f(x_0)-p_0x_0 = -f^\star(p)$ Thus, the tangent line has slope $$p$$ and $$y$$-intercept $$-f^\star(p)$$, which gives a way of visualizing the Legendre transform. Also, this shows the projective duality, in which $$y=f(x)$$ gives a function on the Euclidean line $$\RR$$, and $$f^\star(p)$$ gives a mapping on the set of lines $$y=px$$, in other words, the projective line.

If $$f:U\subseteq\RR^m\rightarrow\RR$$ then $$f^\star$$ is defined on the range of $$\nabla f$$ (also a subset of $$\RR^m$$) by $f^\star(y) + f(x) = \langle y,x \rangle \text{ where } y = \nabla f(x)$

Let $$f(x)=x^2$$. Then $$p=f'(x)=2x$$ and so $$x=\frac{1}{2}p$$. Then $f^\star(p) = px - f(x) = \frac{1}{2}p^2 - \frac{1}{4}p^2 = \frac{1}{4}p^2$

## The Hamiltonian

The Hamiltonian $$H$$ is defined in terms of the generalized coordinates $$q_i$$ and the generalized momenta $$p_i$$ defined by $p_i = \frac{\d L}{\d\dot{q}_i}$ Since the Lagrangian $$L$$ is expressed in terms of $$q_i$$ and $$\dot{q}_i$$ and the Hamiltonian seeks to express the equations of motion in terms of $$q_i$$ and $$\d L/\d \dot{q}_i$$, the Hamiltonian is the Legendre Transform of the Lagrangian (with respect to $$(\dot{q}_1,\ldots,\dot{q}_n)$$): $H(\mathbf{q}, \mathbf{p}, t) = \sum_{i=1}^n p_i\dot{q}_i - L(\mathbf{q}, \dot{\mathbf{q}}, t)$ where the RHS needs to be rewritten in terms of $$q_i$$ and $$p_i$$.

In the case when the generalized coordinates are simply Cartesian coordinates and $$V$$ depends only on $$\mathbf{q}$$ (not on $$\dot{\mathbf{q}}$$), then $p_i = \frac{\d L}{\d \dot{q_i}} = \frac{\d}{\d \dot{q_i}}\sum\frac{1}{2} m \dot{q}_j^2 = m\dot{q}_i$ and the generalized momenta are just the ordinary components of momentum. In this case $$\sum_{i=1}^n p_i\dot{q}_i$$ is equal to twice the kinetic energy, $$T$$, and then $$H=2T-L=2T-(T-V)=T+V$$ is the total energy of the system.

In quantization, it is the _generalized momenta, $$q_i$$, which are quantized as $$-i\hbar d/dx$$, even when they don't align with the momentum in Cartesian coordinates.

Hamilton's equations are derived by calculating the derivative of $$H$$ and using the definition of $$p_i=\d L/\d\dot{q}_i$$ together with Lagrange's Equations, which become $$\d L/\d q_i = \dot{p}_i$$: $\begin{split} dH &= \sum \left( \dot{q}_i dp_i + p_i d\dot{q}_i - \frac{\d L}{\d q_i}dq_i - \frac{\d L}{\d\dot{q}_i}d\dot{q}_i \right) - \frac{\d L}{\d t}dt \\ &= \sum\left( \dot{q}_i dp_i + p_i d\dot{q}_i - \dot{p}_i dq_i - p_i d\dot{q}_i \right) - \frac{\d L}{\d t}dt \\ &= \sum\left( \dot{q}_i dp_i - \dot{p}_i dq_i \right) - \frac{\d L}{\d t}dt \end{split}$ On the other hand, from first principles, $dH = \sum\left( \frac{\d H}{\d p_i}dp_i + \frac{\d H}{\d q_i}dq_i \right) + \frac{\d H}{\d t}dt$ and so, comparing coefficients, one obtains Hamilton's Equations: $\begin{split} \frac{\d H}{\d p_i} &= \dot{q}_i \\ \frac{\d H}{\d q_i} &= -\dot{p}_i\\ \frac{\d H}{\d t} &= - \frac{\d L}{\d t} \end{split}$

A charged particle of charge $$e$$ is moving in an electric field $$\mathbf{E}$$ and magnetic field $$\mathbf{B}$$. From two of Maxwerll's Laws $\nabla\cdot\mathbf{B} = 0 \quad\text{and}\quad \nabla\times\mathbf{R} = -\frac{\d \mathbf{B}}{\d t}$ Thus there is a vector field $$\mathbf{A}$$ and a scalr field $$\phi$$ such that $\mathbf{B} = \nabla\times\mathbf{A} \quad\text{and}\quad \mathbf{E} = -\nabla\phi - \frac{\d\mathbf{A}}{\d t}$ The Lagrangian for this system (in Cartesian coordinates) is $L = \frac{1}{2}m \dot{\mathbf{q}}^2 + e \mathbf{A}\cdot\dot{\mathbf{q}} - e\phi$ (I don't really know why this is -- I want to understand this better!!) So in this case the canonical momenta $$p_i$$ are $p_i = \frac{\d L}{\d \dot{q_i}} = m\dot{q_i} + e A_i$ and so $m\dot{\mathbf{q}} = \mathbf{p} - e\mathbf{A}$ Thus the Hamiltonian is \begin{align} H &= \mathbf{p}\cdot\dot{\mathbf{q}} - L \\ &= \frac{1}{m} \mathbf{p}\cdot(\mathbf{p} - e\mathbf{A}) - \frac{1}{2m}(\mathbf{p} - e\mathbf{A})^2 - \frac{e}{m}\mathbf{A}\cdot(\mathbf{p} - e\mathbf{A}) + e\phi \\ &= \frac{1}{2m}(\mathbf{p} - e\mathbf{A}) \cdot [2\mathbf{p} - (\mathbf{p} - e\mathbf{A}) - 2e\mathbf{A}] + e\phi \\ &= \frac{1}{2m}(\mathbf{p} - e\mathbf{A})^2 + e\phi \end{align} The quantized version of this (obtained by quantizing the canonical momenta) is $i\hbar\frac{d\Psi}{dt} = \hat{H}\psi$ where $\hat{H}\Psi = \left[\frac{1}{2m}\left( -i\hbar\frac{d}{dx} - e \mathbf{A} \right)^2 + e\phi\right]\Psi$